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\title{CHAPTER 14: INVERSE IMAGES}
\author{SCC ET AL}
%\institute[XX大学]{XX大学\quad 数学与统计学院\quad 数学与应用数学专业}
%\date{2025年6月}

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% 封面页
\begin{frame}
  \titlepage
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% 目录页
\begin{frame}{Contents}
  \tableofcontents
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% Section 0
%\section{INTRO.}
\begin{frame}{intro. }

In this chapter we show how to change the ring of scalars of a module over a Weyl algebra. 

This is very important and leads to the construction of many new modules. 

As we shall see later, holonomic modules are preserved under this construction.

\end{frame}

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% Section 1
\section{CHANGE OF RINGS}
\begin{frame}[allowframebreaks]{A. }

We start with a general construction for rings and modules. 

Let $R$, $S$ be rings, and let $\phi: R \to S$ be a ring homomorphism. 

If $M$ is a left $R$-module, then we may use $\phi$ to turn it into a left $S$-module. 

This is called {\color{red}changing the base ring}. 

The key to the construction is that $S$ may be considered as a right $R$-module in the following way. 

Let $r \in R$, $s \in S$; the right action of $R$ on $S$ is defined by
\[ s * r = s \phi(r). \]

The juxtaposition on the right hand side of the equation denotes multiplication in the ring $S$.

With this in mind, we may consider $S$ as an $S$-$R$-bimodule. 

We are now allowed to take the tensor product $S \otimes_R M$ which is a left $S$-module. 

Thus starting with the left $R$-module $M$ we have constructed the left $S$-module $S \otimes_R M$. 

If it is necessary to call attention to the homomorphism $\phi$, we will write $S \otimes_\phi M$.

As an application, let us rephrase the twisting construction of Ch. 5, §2 in terms of change of rings.

\textbf{1.1 LEMMA.} Let $R$ be a ring and $\sigma$ an automorphism of $R$. 

Let $M$ be a left $R$-module. Then
\[ M_\sigma \cong R \otimes_{\sigma^{-1}} M. \]

\textbf{PROOF:} Let $a \in R$. 

The left action on $M_\sigma$ is given by $a \bullet u = \sigma(a)u$, where $u \in M_\sigma$; and the right action on $R$ by $s * a = s \sigma^{-1}(a)$, where $s \in R$. 

Consider the map
\[ \phi: R \times M \to M_\sigma \]
defined by $\phi(a, u) = a \bullet u$. 

Since
\[
\phi(a, bu) = a \bullet bu = (\sigma(a)b)u
\]
is equal to $(a\sigma^{-1}(b)) \bullet u = \phi(a*b, u)$, the map is balanced. 

It is also bilinear. 

Thus, by the universal property of the tensor product, there exists a homomorphism
\[
\overline{\phi}: R \otimes_{\sigma^{-1}} M \to M_\sigma.
\]

Since $M_\sigma = M$ as abelian groups, the map $\overline{\phi}$ is surjective.

Let $v \in R \otimes_{\sigma^{-1}} M$. There exists $u \in M$ such that $v = 1 \otimes u$. 

Hence
\[
0 = \overline{\phi}(v) = u,
\]
which implies that $u = 0$. 

Thus $v = 0$ and $\overline{\phi}$ is injective. 

Therefore, $\overline{\phi}$ is an isomorphism of $R$-modules.

We shall now apply the change of rings construction to polynomial maps. 

We often use the results of Ch. 4, §1 in the coming chapters, so you may wish to read that section again before you proceed. 

A good notation will be of great help: we shall abide by the conventions we are about to make right to the end of the book.

\textbf{1.2 NOTATION.} Put $X = K^n$. 

The polynomial ring $K[x_1, \ldots, x_n]$ will be denoted by $K[X]$; and the Weyl algebra generated by the $x$'s and $\partial_x$'s by $A_n$. 

The $n$-tuple $(x_1, \ldots, x_n)$ will be denoted by $X$. 

Similar conventions will hold for $Y = K^m$ and $Z = K^r$, with polynomial rings $K[Y]$ and $K[Z]$ and Weyl algebras $A_m$ and $A_r$. 

The space $K^{m+n}$ equals the cartesian product of $X$ and $Y$ and will be denoted by $X \times Y$. 

For the polynomial ring in the $x$'s and $y$'s we shall write $K[X, Y]$. 

The $(n+m)$-th Weyl algebra $A_{m+n}$ will always denote the algebra generated by $K[X, Y]$, $\partial_x$'s and $\partial_y$'s. 

Note that with these conventions $A_{m+n} = A_m \widehat{\otimes} A_n$.

Although these conventions may cause some initial confusion, they pay off later on by making the formulae easier to digest than they would otherwise be.

Let $F: X \to Y$ be a polynomial map. 

Its comorphism $F{\,}^\sharp: K[Y] \to K[X]$ is an algebra homomorphism. 

If $M$ is a $K[Y]$-module, we may use $F{\,}^\sharp$ to construct the $K[X]$-module $K[X] \otimes_{K[Y]} M$. 

It is called the {\color{red}inverse image} of $M$ by the polynomial map $F$, and denoted by $F{\,}^*M$. 

Let us consider some examples.

Let $F: X \to X$ be a polynomial isomorphism, with inverse $G$. 

The comorphism $F{\,}^\sharp$ is an automorphism of $K[X]$ with inverse $G{\,}^\sharp$. 

Let $M$ be a $K[X]$-module. 

By Lemma 1.1, its {\color{red}inverse image} $F{\,}^*M$ is isomorphic to $M_\sigma$. 

Consequently, if $u \in F{\,}^*M$ and $h \in K[X]$, then $hu = G{\,}^\sharp(h)u$.

The second example is the projection $\pi: X \times Y \to Y$ onto the second coordinate: $\pi(X,Y) = Y$. 

The comorphism $\pi{\,}^\sharp$ maps a polynomial $p \in K[Y]$ to itself, but considered as an element of $K[X,Y]$. 

Let $M$ be a module over $K[Y]$, then
\[
\pi^*M = K[X,Y] \otimes_{K[Y]} M.
\]

A monomial in $K[X,Y]$ may be written in the form $pq$, with $p \in K[X]$ and $q \in K[Y]$. 

If $u \in M$ then
\(
pq \otimes u = p \otimes qu
\)
as elements of $\pi^*M$. 

Using this identity we may construct an isomorphism,
\[
\pi^*M \cong K[X] \widehat{\otimes} M,
\]
of $K[X,Y]$-modules.

\end{frame}

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% Section 2
\section{INVERSE IMAGES.}
\begin{frame}[allowframebreaks]{B. }

Let $F: X \to Y$ be a polynomial map and $M$ a left $A_m$-module. 

Since $Y = K^m$, it follows that $M$ is in particular a $K[Y]$-module. 

Thus we may compute the {\color{red}inverse image} of $M$ by $F$ defined in §1:
\[
F{\,}^*(M) = K[X] \otimes_{K[Y]} M
\]
is a module over $K[X]$. 

We want to make this module into an $A_n$-module.

At first sight this construction may seem very puzzling. 

Why do {\color{red}change of rings} as if we had only a module over a polynomial ring and then turn it back into a module over a Weyl algebra? 

Why not do the change of rings at the Weyl algebra level? 

First of all, one is allowed to do change of rings at the level of Weyl algebras, since the construction works for any ring. 

However, in doing that one is limited by the fact that Weyl algebras are simple rings; and homomorphisms of simple rings are necessarily injective. 

A more satisfactory answer is that this is really a construction in algebraic geometry, that we are lifting to the realm of noncommutative rings; see [Hartshorne 77, Ch. 2, §5]. 

Anyhow, the efficacy of the construction is its best justification, as we shall see later on.

We know how a polynomial of $K[X]$ acts on $F{\,}^*(M)$. 

We now give the recipe for the action of $\partial_{x_i}$. 

Let
\[ q \otimes u \in F{\,}^*(M), \]
where $q \in K[X]$ and $u \in M$. 

Let $F_1, \ldots, F_m$ be the coordinate functions of $F$. 

The action of $\partial_{x_i}$ is defined by
\begin{equation}
    \partial_{x_i}(q \otimes u) = \partial_{x_i}(q) \otimes u + \sum_{k=1}^{m} q \partial_{x_i}(F_k) \otimes \partial_{y_k} u
\end{equation}
for $i = 1, \ldots, n$.

The $x$'s and $\partial_x$'s generate $A_n$, and we know how they act on $F{\,}^*(M)$. 

According to Appendix 1, these formulae will make an $A_n$-module of $F{\,}^*(M)$ if they are compatible with the relations satisfied by the generators. 

The relations are
\begin{align*}
    [\partial_{x_i}, x_j] &= \delta_{ij} 1, \\
    [x_i, x_j] &= [\partial_{x_i}, \partial_{x_j}] = 0,
\end{align*}
for $1 \leq i,j \leq n$. 

Let us carefully check that the first of these relations is compatible with the actions of $x$'s and $\partial_x$'s defined above.

Let $w \in F{\,}^*(M)$. We want to show that
\[ [\partial_{x_j}, x_i] w = \delta_{ij} w. \]

Clearly it is enough to check this when $w$ is of the form $q \otimes u$, for $q \in K[X]$ and $u \in M$. 

Let us begin by calculating $\partial_{x_j}(x_i(q \otimes u))$. 

We do this by applying (2.1) to $x_i q \otimes u$, which gives
\begin{equation}
    \partial_{x_j}(x_i q) \otimes u + \sum_{1}^{m} x_i q \partial_{x_j}(F_k) \otimes \partial_{y_k} u.
\end{equation}


Using Leibniz's rule and the fact that $\partial_{z_j}(x_i) = \delta_{ij}$, we may rewrite (2.2) in the form
\[
\delta_{ij} q \otimes u + x_i \left( \partial_{z_j}(q) \otimes u + \sum_{1}^{m} (q \partial_{z_j}(F_k) \otimes \partial_{y_k} u) \right).
\]

Note that the expression in brackets equals $\partial_{z_j}(q \otimes u)$. 

Thus we have shown that
\[
\partial_{z_j}(x_i(q \otimes u)) = \delta_{ij} q \otimes u + x_i \partial_{z_j}(q \otimes u).
\]

But this is equivalent to
\[
[\partial_{z_j}, x_i](q \otimes u) = \delta_{ij}(q \otimes u),
\]
which is what we wanted to prove.

The proof that the other relations are compatible with these actions is an exercise for the reader. 

We will calculate some concrete examples in the next section.

\end{frame}

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% Section 3
\section{PROJECTIONS.}
\begin{frame}[allowframebreaks]{C. }

Let us return to the projection $\pi: X \times Y \to Y$ of §1. 

Let $M$ be a left $A_m$-module. 

We have already seen that, as a $K[X,Y]$-module, the {\color{red}inverse image} $\pi^*M$ is isomorphic to $K[X] \widehat{\otimes} M$. 

However, $K[X]$ is an $A_n$-module and $M$ is an $A_m$-module. 

So $K[X] \widehat{\otimes} M$ has a natural $A_{m+n}$-module structure. 

We want to show that this structure coincides with the one determined by (2.1).

It is enough to check how $\partial_{x_i}$ and $\partial_{y_j}$ act on $K[X] \widehat{\otimes} M$. 

For this we return to the definitions. 

The $K[X,Y]$-module isomorphism between $\pi^*(M)$ and $K[X] \widehat{\otimes} M$ maps $q_\alpha x^\alpha \otimes u$ to $x^\alpha \otimes q_\alpha u$, where $\alpha \in \mathbb{N}^n$ and $q_\alpha \in K[Y]$. 

We must show that this is compatible with the action of the derivatives.

Consider first the action of $\partial_{y_j}$, as defined by (2.1),
\[
\partial_{y_j}(q_\alpha x^\alpha \otimes u) = \partial_{y_j}(q_\alpha x^\alpha) \otimes u + \sum_{1}^{m} (\partial_{y_j}(y_k) q_\alpha x^\alpha \otimes \partial_{y_k} u).
\]

Since $\partial_{y_j}(x^\alpha) = 0$ and $\partial_{y_j}(y_k) = \delta_{jk}$, we obtain
\[
\partial_{y_j}(q_\alpha x^\alpha \otimes u) = x^\alpha \partial_{y_j}(q_\alpha) \otimes u + q_\alpha x^\alpha \otimes \partial_{y_j} u.
\]


Using Leibniz's rule, this may be rewritten as
\[
\partial_y (q_\alpha x^\alpha \otimes u) = x^\alpha \otimes \partial_y (q_\alpha u).
\]

Summing up: if we identify $\pi^*(M)$ with $K[X] \widehat{\otimes} M$, then $\partial_y$ acts only on $M$, in the usual way.

Consider now the action of $\partial_{x_i}$. 

From the definition, we have that
\[
\partial_{x_i}(q_\alpha x^\alpha \otimes u) = \partial_{x_i}(q_\alpha x^\alpha) \otimes u + \sum_{1}^{m} \partial_{x_i}(y_k) q_\alpha x^\alpha \otimes \partial_{x_k} u.
\]

Since $\partial_{x_i}(y_k) = \partial_{x_i}(q_\alpha) = 0$, for all $k$ and all $\alpha$, we end up with
\[
\partial_{x_i}(q_\alpha x^\alpha \otimes u) = q_\alpha \partial_{x_i}(x^\alpha) \otimes u.
\]

Thus, under the previous identification, $\partial_{x_i}$ acts only on $K[X]$.

We may sum up our calculations as follows. 

If the module $\pi^*(M)$ is identified with $K[X] \widehat{\otimes} M$ then the $x$'s and $\partial_x$'s act only on the first factor $K[X]$, and the $y$'s and $\partial_y$'s act only on the second factor $M$. 

Besides, the actions are the natural ones.

We may use this description to calculate the dimension of an {\color{red}inverse image}.

\textbf{3.1 THEOREM.} Let $M$ be a finitely generated left $A_m$-module and $\pi$ the projection defined above. Then:
\begin{enumerate}
    \item[(1)] $\pi^*M$ is a finitely generated $A_{m+n}$-module.
    \item[(2)] $d(\pi^*M) = n + d(M)$.
    \item[(3)] $m(\pi^*M) \leq m(M)$.
\end{enumerate}

\textbf{PROOF:} (1) follows from Proposition 13.2.1; whilst (2) and (3) are consequences of Theorem 13.4.1 since, as an $A_n$-module, $K[X]$ has dimension $n$ and multiplicity 1; see Ch. 9, §2.

\textbf{3.2 COROLLARY.} Let $M$ be a holonomic $A_m$-module. Then $\pi^*M$ is a holonomic $A_{m+n}$-module.

\end{frame}

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% Section 4
\section{4. EXERCISES}
\begin{frame}[allowframebreaks]{D. }

\begin{enumerate}\itemsep2em

\item  Exercise {4.1} Let $R$, $S$, $T$ be rings and $\phi: R \to S$ and $\psi: S \to T$ be ring homomorphisms. 

If $M$ is a left $R$-module, show that
\[
T \otimes_\psi S \otimes_\phi M \cong T \otimes_{\psi \circ \phi} M.
\]

\newpage

\item  Exercise {4.2} Let $F: K \to K$ be the polynomial map defined by $F(x) = x^2$. 

Let $M$ be the algebra $A_1$ considered as a left module over itself. 

Show that $F{\,}^*M$ is not finitely generated over $A_1$.

\newpage

\item  Exercise {4.3} Let $\sigma$ be an automorphism of $A_n$. 

Suppose that the restriction $\theta$ of $\sigma$ to $K[X]$ is an automorphism of this ring. 

Put $F = \theta_\sharp$ for the polynomial map determined by $\theta$. 

Let $M$ be a left $A_n$-module. 

Is it true that
\[ F{\,}^*M \cong M_\sigma \]
as $A_n$-modules?

\newpage

\item  Exercise {4.4} Let $M$ and $N$ be left $A_n$-modules. 

In particular these are modules over $K[X]$. 

Hence the tensor product $M \otimes_{K[X]} N$ is a well-defined $K[X]$-module. 

Define the action of $\partial_{x_i}$ on $u \otimes v \in M \otimes_{K[X]} N$ by the formula
\[ \partial_{x_i}(u \otimes v) = \partial_{x_i} u \otimes v + u \otimes \partial_{x_i} v. \]

\begin{enumerate}
    \item[(1)] Proceeding as in §2, show that this action makes $M \otimes_{K[X]} N$ into an $A_n$-module.
    \item[(2)] Give an example of two finitely generated $A_n$-modules $M$ and $N$ such that $M \otimes_{K[X]} N$ is not finitely generated over $A_n$.
\end{enumerate}

\newpage

\item  Exercise {4.5} Let $F: \mathbb{C} \to \mathbb{C}$ be the map defined by $y = F(x) = x^m$, where $m \geq 2$ is an integer. 

Let $\delta$ be the Dirac microfunction; see Ch. 6, §3. 

The purpose of this exercise is to show that the {\color{red}inverse image} $F{\,}^*(A_1(\mathbb{C})\delta)$ is isomorphic to the $A_1(\mathbb{C})$-module generated by $\delta^m$, the $m$-th derivative of $\delta$.

\begin{enumerate}
    \item[(1)] Show that $x^m$ and $x \partial_x + m$ annihilate $1 \otimes \delta$.
    \item[(2)] Show by induction that there exist non-zero complex numbers $c_{pq}$ such that
    \[ \partial_x^{mp+q}(1 \otimes \delta) = c_{pq} x^{m-q} \otimes \partial_y^{p+1} \delta, \]
    for $q = 0, 1, \ldots, m$.
    \item[(3)] Using (2) show that
    \[ 1 \otimes \partial_y^{p+1} \delta = \frac{1}{c_{pm}} \partial_x^{(p+1)m}(1 \otimes \delta). \]
    \item[(4)] Conclude from (3) that $1 \otimes \delta$ generates $F{\,}^*(A_1(\mathbb{C})\delta)$.
    \item[(5)] By (1), (3) and Exercise 6.4.10, $F{\,}^*(A_1(\mathbb{C})\delta)$ is a homomorphic image of $A_1(\mathbb{C})\delta^m$. Since the latter is irreducible, we have the desired isomorphism.
\end{enumerate}

In the language of microfunctions we have proved that the {\color{red}inverse image} of $\delta$ under $F$ is $\delta^m$.

\end{enumerate}

\end{frame}
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